Basis Graphs of Even Delta-Matroids

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Journal of Combinatorial Theory, Series B ••• (••••) •••–••• www.elsevier.com/locate/jctb

Basis graphs of even Delta-matroids

Victor Chepoi

Laboratoire d’Informatique Fondamentale de Marseille, Faculté des Sciences de Luminy, Université de la Méditerranée, F-13288 Marseille cedex 9, France Received 16 February 2000

Abstract A -matroid is a collection B of subsets of a ﬁnite set I, called bases, not necessarily equicardinal, satisfying the symmetric exchange property: For A,B ∈ B and i ∈ AB, there exists j ∈ BAsuch that (A {i, j}) ∈ B.A-matroid whose bases all have the same cardinality modulo 2 is called an even -matroid.Thebasis graph G = G(B) of an even -matroid B is the graph whose vertices are the bases of B and edges are the pairs A,B of bases differing by a single exchange (i.e., |AB|=2). In this note, we present a characterization of basis graphs of even -matroids, extending the description of basis graphs of ordinary matroids given by S. Maurer in 1973: Theorem. A graph G = (V, E) is a basis graph of an even -matroid if and only if it satisﬁes the following conditions:

(a) if x1x2x3x4 is a square and b ∈ V ,thend(b,x1) + d(b,x3) = d(b,x2) + d(b,x4); (b) each 2-interval of G contains a square and is an induced subgraph of the 4-octahedron; (c) the neighborhoods of vertices induce line graphs, or, equivalently, the neighborhoods of vertices do not contain induced 5- and 6-wheels. (A 2-interval is the subgraph induced by two vertices at distance 2 and all their common neighbors; a square is an induced 4-cycle of G.) © 2006 Elsevier Inc. All rights reserved.

Keywords: Basis graph; -Matroid; Graph distance

E-mail address: [email protected].

0095-8956/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jctb.2006.05.003 ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.2 (1-18) 2 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

1. Introduction

Matroids constitute an important unifying structure in combinatorics, algorithmics, and combinatorial optimization. According to one of the many equivalent deﬁnitions, a matroid on a ﬁnite set of elements I is a collection B of subsets of I , called bases, which satisfy the following exchange property:

(EP) For all A,B ∈ B and i ∈ A \ B there exists j ∈ B \ A such that A \{i}∪{j}∈B.

We say that the base A \{i}∪{j} is obtained from the base A by an elementary exchange.It is well known that all the bases of a matroid have the same cardinality, which is called its rank. The basis graph G = G(B) of a matroid B is the graph whose vertices are the bases of B and edges are the pairs A,B of bases differing by a single exchange (i.e., |AB|=2, where the symmetric difference of two sets A and B is written and defined by AB = (A\B)∪(B \A)). It has been shown by Cunningham (unpublished), Holzmann, Norton, and Tobey [18], and Maurer [19] that basis graphs faithfully represent their matroids, thus studying the basis graph amounts to studying the matroid itself (see [19] for some further references on basis graphs). Gelfand, Goresky, MacPherson, and Serganova [17] defined a basis matroid polyhedron as the convex hull of characteristic vectors of bases of a matroid. They showed that a convex hull of characteristic vectors of a collection A of equicardinal subsets of an n-element set is a basis matroid polytope if and only if its 1-skeleton is isomorphic to the basis graph of the family A. In [19], Maurer presented a characterization of graphs which are basis graphs of matroids and described basis graphs of several important classes of matroids, in particular, of binary matroids (see also [14] for a characterization of basis graphs of uniform matroids and [20] for investigation of properties of intervals of matroid basis graphs). From this characterization easily follows that all basis graphs are homotopically trivial, a property used several times in the theory of ordinary and oriented matroids; cf. [2,3,16]. From this result also follows that the 2-dimensional faces of a basis matroid polytope are equilateral triangles or squares; cf. [6]. There are several important and interesting generalizations of the concept of matroid. Delta- matroids is one of them, and have been introduced independently by Bouchet [9–11], Chan- drasekaran and Kabadi [13], and Dress and Havel [15] in essentially equivalent ways. A - matroid is a collection B of subsets of a finite set I , called bases, not necessarily equicardinal, satisfying the following symmetric exchange property:

(SEP) For A,B ∈ B and i ∈ AB,there exists j ∈ BAsuch that (A {i, j}) ∈ B.

It is immediately clear that the family of bases of a matroid is also a -matroid. In fact, matroids are precisely the -matroids for which all members of B have the same cardinality. A -matroid whose bases all have the same cardinality modulo 2 is called an even -matroid. If A,B are two bases of an even -matroid B and B = A{i, j} we say that B is obtained from A by an elementary exchange. Following the terminology for ordinary matroids, the basis graph G = G(B) of an even -matroid B is the graph whose vertices are the bases of B and edges are the pairs A,B of bases differing by a single exchange, i.e., A and B are adjacent if and only if |AB|=2. Some properties of these graphs have been used and investigated by Wenzel [22,23]. For a subset I of I denote B I := {BI: B ∈ B} and say that B I is obtained by applying a twisting to B. Then B I is a -matroid. It can be easily shown that the even - matroids B and B I have isomorphic basis graphs. Applying a twisting to a matroid we will ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.3 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 3 always obtain an even -matroid, but only in some cases this will be a matroid (for example, if I = I, we will get the dual matroid). It will be convenient to identify I with the set [n]={1, 2,...,n}. Let I ∗ ={1∗, 2∗,...,n∗}. Bouchet deﬁnes a symmetric matroid as essentially a -matroid with bases extended to n elements by adding to B ∈ B all starred elements from I ∗ which do not appear, unstarred, in B. Thus a symmetric matroid is a family B of subsets of cardinality n of the set I ∪ I ∗ such that in every subset of B each element of I appears either unstarred or starred and if

For A,B ∈ B and i ∈ AB, there exists j ∈ BAsuch that (A {i, j, i∗,j∗}) ∈ B.

Many important properties associated with matroids (greedy algorithm, polyhedral description) extend to -matroids. Apart from well-known examples of matroids, several other discrete structures satisfy (SEP). For example, consider a skew-symmetric matrix M = (mij : i, j ∈ I) (i.e., M = (−M)T and all diagonal entries of M are zero) and define B by letting B ∈ B if and only if the principal submatrix (mij : i, j ∈ B) is non-singular. Then B is an even -matroid [10]. Second, let G = (V, E) be a graph, and a subset of vertices B of G belongs to B if and only if there is a matching of G covering precisely the elements of B. The resulting collection B is an even -matroid [11,13]. Another nice instance of an even -matroid is given in [4,12] and arises by considering a graph G = (V, E) drawn on a compact surface S and its geometric dual G∗ = (F, E∗): e∗ ∈ E∗ is the unique edge of G∗ which cuts the edge e ∈ E. Let B consist of maximal by inclusion subsets B of E ∪ E∗ (e and e∗ cannot be simultaneously in B) such that the surface S is not disconnected by cutting it along the edges from B. Then B is an even -matroid. Finally notice that -matroids occupy an important place in the theory of Coxeter matroids (where one can find them under the name of “Lagrangian matroids”) [5,7,8]. In this note, we characterize the basis graphs of even -matroids, extending and refining Maurer’s description of basis graphs of matroids. Our proof provides an alternative approach to Maurer’s result: departing from a graph satisfying Maurer’s conditions and performing a suitable twisting to the even -matroid obtained in our proof, we get a matroid.

2. Terminology and main results

In this section we establish our notation and formulate the principal results. All graphs G = (V, E) occurring here are ﬁnite, connected, and without loops or multiple edges. For brevi- ty’s sake, we use the notation x ∼ y if x,y are adjacent and x y otherwise. If x ∼ y, we denote the corresponding edge by xy. By a square x1x2x3x4 we will mean an induced 4-cycle with x1 x3 and x2 x4. The distance d(u,v) := dG(u, v) between two vertices u and v of a graph G = (V, E) is the length of a shortest path between u and v. The set of all vertices w on shortest (u, v)-paths is the interval [u, v]. For convenience, we will use the short-hand (u, v) =[u, v]\{u, v} to denote the “interior” of the interval between u and v. If d(u,v) = 2, we say that [u, v] is a 2-interval. A subset of vertices S of G (or the subgraph G(S) induced by S) is called convex if [u, v]⊆S for any u, v ∈ S. The convex hull conv(A)ofasetA is the smallest convex set containing A. An induced subgraph H of G is isometric if the distance between any pair of vertices in H is the same as that in G. It will be convenient to use the same notation and terminology for a subset of vertices and the subgraph induced by this subset; for example, [u, v] will also denote the subgraph induced by the interval between u and v. The Cartesian product G = G1 G2 of two graphs G1 and G2 has the pairs (x1,x2) as its vertices (with vertex xi from Gi ) and an edge between two vertices x = (x1,x2) and y = (y1,y2) ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.4 (1-18) 4 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• if and only if, for some i ∈{1, 2}, the vertices xi and yi are adjacent in Gi, and xj = yj for the remaining j = i. The sum of two even -matroids B1 and B2 deﬁned on disjoint ground sets is the even -matroid B1 + B2 ={B1 ∪ B2: B1 ∈ B1,B2 ∈ B2}. In that case, we obtain G(B1 + B2) = G(B1) G(B2). Three important classes of graphs serve as host spaces for basis graphs in question. Denote by =| | ∈ [n] [ ] Hn the n-cube; notice that dHn (A, B) AB for any A,B 2 . Let A,B Hn be the interval [ ] ={ ∩ ⊆ ⊆ ∪ } 1 between A and B; clearly A,B Hn C: A B C A B . The half-cube 2 Hn is the graph whose vertex set is the collection of all subsets of [n] which have the same cardinality modulo 2 1 | |= 1 and two vertices A,B are adjacent in 2 Hn iff AB 2. Given two vertices A,B of 2 Hn, we 1 denote by [A,B] 1 all subsets of [A,B] which are vertices of H . The Johnson graph J Hn Hn 2 n n,k 2 1 is the (isometric) subgraph of 2 Hn induced by the family of all subsets of cardinality k. Axioms (EP) and (SEP) imply that the basis graphs of matroids of rank k are isometric subgraphs of 1 Jn,k, while the basis graphs of even -matroids are isometric subgraphs of 2 Hn. Recall also that the m-octahedron Km×2 is the complete multipartite graph with m 2 parts, each of size 2. Equivalently, Km×2 is obtained from the complete graph K2m by deleting a perfect matching. 1 Notice that all 2-intervals in the half-cube 2 Hn with n 4 are 4-octahedra and all 2-intervals in the Johnson graph Jn,k with n 4 and 1

(ICm) Every 2-interval [u, v] contains a square K2×2 and is an induced subgraph of the m- octahedron Km×2.

A levelling of a graph G = (V, E) from a base-point b is a partition of V into the sets Ni(b) = {v ∈ V : d(b,v) = i}. We will denote by N(b) the subgraph induced by N1(b). Define a partial order ≺b on V by letting u ≺b v iff d(b,v) = d(b,u) + d(u,v). Maurer [19] established that a graph G is a matroid basis graph if and only if it satisfies (i) the interval condition (IC3), (ii) the neighborhood N(b) of some vertex b is the line graph of a bipartite graph, and (iii) in any levelling of G each octahedral 2-interval lies in one of three positions: (1) all in one level; (2) in two levels, three adjacent vertices in each; or (3) in three levels, a square in between, one vertex in the highest, and one vertex in the lowest, any other 2-interval lies as an induced subgraph of a 3-octahedron positioned as above. We will use another (apparently simpler) positioning condition, which, as we will show in Section 4, it is satisfied by half-cubes and all their isometric subgraphs:

(PC) For every vertex b and every square v1v2v3v4 of G, the following equality holds:

d(b,v1) + d(b,v3) = d(b,v2) + d(b,v4).

We will call a square upward if it belongs to three consecutive levels, horizontal if it lies in one level, and vertical if two adjacent vertices are in one level and other two adjacent vertices are in the next level. An edge uv of G is horizontal if u and v lie in one level and upward otherwise. A k-wheel Wk is the graph consisting of a k-cycle plus an additional vertex adjacent to all vertices ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.5 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 5 of the cycle. Finally, recall that the line graph L(Γ ) of a graph Γ has the edges of Γ as vertices and two vertices are adjacent in L(Γ ) if and only if the corresponding edges are incident in Γ . We are ready to formulate the main results of this note.

Theorem 1. For a graph G = (V, E) the following conditions are equivalent:

(i) G is a basis graph of an even -matroid; (ii) G satisﬁes the interval condition (IC4), the positioning condition (PC), and no neighborhood N(v) contains an induced 5-wheel W5 or 6-wheel W6; (iii) G satisﬁes the interval condition (IC4), the positioning condition (PC), and the neighborhood of every vertex is a line graph; (iv) G is a connected induced subgraph of a half-cube satisfying the interval condition (IC4); (v) G is an isometric subgraph of a half-cube satisfying the interval condition (IC4).

The proof of Theorem 1 uses several auxiliary results given in Section 3 and is presented in Section 4. The central issue of this proof is to establish that (iii) implies (iv). We briefly outline the idea of the encoding employed in this proof. Let G = (V, E) be a graph satisfying (IC4) and (PC), such that for some vertex b, N(b) is the line graph of a graph Γ = (I, F ). To establish that G is an induced subgraph of a half-cube, we define the following mapping ϕ : V → 2I . Set ϕ(b) =∅. Each vertexx ∈ N(b) encodes some edge ij of Γ ; put ϕ(x) ={i, j}. For any other vertex v, let ϕ(v) = {ϕ(x): x ∈[b,v]∩N(b)}. We show that ϕ is injective and that all sets ϕ(v) have even cardinality. Finally, we show that ϕ is an edge-preserving map from G to the 1 ⇒ ⇒ ⇒ half-cube 2 Hn, thus establishing (iii) (iv). From the proof of (iv) (v) (i), we conclude that the resulting collection Bϕ := {ϕ(v): v ∈ V } is an even -matroid whose basis graph is G. The proof of the implication (ii) ⇒ (iii) is based on the case analysis of forbidden subgraphs of line graphs and is postponed to the final Section 7. We show that, if Γ is a bipartite graph with I = A ∪ B, then Bϕ B is a matroid of rank |B|, thus providing an alternative proof of Maurer’s characterization. Our encoding scheme is different from that used by Maurer [19], except the encoding of the vertices of N(b), where both schemes are essentially the same. Recall, he encodes the vertex b by B, a vertex x ∈ N(b) repre- senting the edge ij of Γ with i ∈ B and j ∈ A is labelled by the set B \{i}∪{j}. The encoding is inductively expanded to the whole graph using certain upward squares (among other things, in establishing that this labelling is well defined, it is necessary to show that it is independent of chosen squares); see [19] for all details. Notice also that for matroids, a result similar to the equivalence (i) ⇔ (iv) of Theorem 1 is given in Theorem 2.2 of [19]. In Section 5, we establish the following decomposition result of basis graphs:

Proposition 1. Let G be a basis graph of an even -matroid such that the neighborhood N(b) of some vertex b is not connected. If N(b) is the line graph of a graph Γ = (I, F ) and (I1,F1) is a connected component of Γ, then Bϕ is the sum of the even -matroids B1 ={B ∈ Bϕ: B ⊆ I1} and B2 ={B ∈ Bϕ: B ⊆ I \ I1}, and consequently G = G(B1) G(B2).

In view of this result, we call a basis graph indecomposable if the neighborhoods of all its vertices induce connected graphs. Deﬁne the support of an even -matroid to be the elements of the ground set that actually appear in at least one base. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.6 (1-18) 6 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

Proposition 2. Let B1 and B2 be two even -matroids with supports I1 and I2, and isomorphic indecomposable basis graphs G(B1) G(B2) G. Then there exists a bijection α : I1 → I2 and a set S ⊆ I2 so that B2 ={α(B) S: B ∈ B1}, unless G is an induced subgraph of the 4-octahedron.

In fact, our proof yields something a little stronger. Let Bi(v) denote the set in Bi represented byavertexv of G, for i = 1or2. Then α and S have the property that B2(v) = α(B1(v)) S for every vertex v of G. In other words, unless the basis graphs are isomorphic to an induced subgraph of the 4-octahedron, every isomorphism between basis graphs (represented by the iden- tiﬁcations of vertices of G(B1) and G(B2) with vertices of G) is induced by a bijection from one support to the other followed by a twisting. In Section 6, we extend the characterization of Gelfand et al. [17] of basis matroid polyhedra to basis polyhedra of even -matroids. Let B ⊂ 2[n]. For B ∈ B, let σ(B) be its incidence 0, 1-vector: σi(B) = 1ifi ∈ B and σi(B) = 0 otherwise. Let Π(B) be the convex hull (in the usual sense) of the points σ(B), B ∈ B. Denote by Π1(B) the 1-skeleton of Π(B): it has the vertices of Π(B) as the set of vertices and two vertices are adjacent in Π1(B) if and only if the (linear) segment connecting them is a 1-dimensional face of the polyhedron Π(B). Notice that B B B 1 for any collection of even subsets , Π1( ) contains the subgraph G( ) of 2 Hn induced by its vertices. Indeed, it is easy to show explicitly that the set E of all even subsets of {1, 2,...,n} 1 B ⊆ B generates a polyhedron having the half-cube 2 Hn as 1-skeleton. Then G( ) Π1( ) follows because any edge of Π1(E) with vertices in B must be an edge of Π1(B) also. As we will show below, the converse inclusion characterizes the even -matroids:

Theorem 2. If B is an even -matroid, then the graph Π1(B) coincides with the basis graph [n] G(B). Conversely, if B ⊂ 2 is a collection of sets of even cardinality and the graphs Π1(B) and G(B) coincide, then B is an even -matroid.

3. Auxiliary results

In this section, we establish some results that will be used throughout the proof of Theorem 1. Unless stated otherwise, G is always a graph fulﬁlling condition (iii) of Theorem 1. For properties (3.1) and (3.2) in case of matroid basis graphs see also [19]. The proofs of (3.3) and (3.4) become much shorter if one replace (IC4) by (IC3).

(3.1) (Triangle condition) If u ∼ v and d(b,u) = d(b,v) = k, then there exists a vertex w ∼ u, v at distance k − 1 to b (a common parent of u, v with respect to ≺b).

Proof. Proceed by induction on k. Pick z ∈[u, b] adjacent to u. If z ∼ v, we are done. Otherwise take a square zxvy. Then x ∈ Nk(b) and y ∈ Nk−1(b) by (PC). If x = u, then u is adjacent to x and y by (IC4), therefore y is the desired parent. Finally, let u = x. By the induction hypothesis, there is a vertex t ∼ z, y at distance k − 2tob. Consider an upward square containing v,t, and two other vertices in Nk−1(b). By (PC), u must be adjacent to one of these vertices. 2

A propeller is a graph obtained by gluing three 3-cycles along a common edge.

(3.2) (No propellers) G does not contain induced propellers. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.7 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 7

Proof. Suppose G has a propeller with tips x1,x2,x3, and the edge y1y2 whose ends are adjacent to all tips. Then N(y2) contains a K1,3 subgraph induced by x1,x2,x3, and y1, which is forbidden in line graphs. 2

(3.3) (Local convexity implies convexity) A connected induced subgraph H of G is convex if and only if whenever x,y ∈ H and d(x,y) = 2, then [x,y]⊆H.

Proof. We will show that [u, v]⊆H for u, v ∈ H by induction on k = dH (u, v), the case dH (u, v) = 2 being covered by the initial assumption. Suppose by way of contradiction that one can ﬁnd u, v ∈ H at distance k 3inH and a vertex w ∈[u, v]\H. Additionally assume that, if there are several such pairs, the selected one has least distance d(u,v) in G. Pick a neighbor z of u on a shortest (u, v)-path P in H . Then [v,z]⊆H by the induction hypothesis, in particular dH (v, z) = d(v,z).Ifd(u,v) = d(z,v), by (3.1) u and z have a common neighbor in [v,z], contrary to the choice of P. Thus d(u,v) = d(z,v) + 1 = dH (u, v).Lett be a neighbor of u inashortest(u, v)-path passing via w.Ift ∼ z, by (3.1) they have a common neighbor s ∈[v,z]⊂H . Since u s, we obtain t ∈[s,u]⊂H . Then, by the choice of u, v, we have [t,v]⊂H , contrary to w ∈[t,v]\H. Thus t z. Consider a square ts1zs2.Ifu is a vertex of this cycle, say u = s2, then d(s1,v)= k − 2 by (PC), yielding s1 ∈ H because s1 ∈[z, v]. Since t ∈[u, s1]⊂H, we obtain the same contradiction as before. Hence u = s1,s2, whence u ∼ s1,s2. Applying (PC) to v and the square s1ts2z we conclude that either d(v,s1) = d(v,s2) = k − 1or d(v,s1) = k − 2 and d(v,s2) = k. In the second case, we could replace s2 by u and have the case where u belongs to the square ts1zs2. Thus d(v,s1) = d(v,s2) = k − 1. If s1,s2 ∈ H, then t ∈[s1,s2]⊂H . By the choice of u, v, each of the intervals [s1,v], [s2,v], and [t,v] belong to H, a contradiction. Thus, let s1 ∈/ H . By (3.1), there exists a vertex p ∼ s1,zat distance k − 2 to v. Since p ∈[z, v]⊂H and s1 ∈[u, p], we get a contradiction with local convexity of H . 2

(3.4) The intervals of G are convex.

Proof. By (3.3) it sufﬁces to establish the local convexity of intervals. Suppose by way of contradiction that there exist u, v ∈ V,x,y ∈[u, v] with d(x,y) = 2, and a vertex z ∈ (x, y)\[u, v].By (PC) all 2-intervals are convex, thus d(u,v) 3. Further, we may assume that [u, x]∩[u, y]= {u}, otherwise u can be replaced by a closest to x and y vertex in this intersection. Analogously, [v,x]∩[v,y]={v}. We distinguish two cases: Case 1. d(u,x) < d(u,y), d(v, x) > d(v, y). Then d(u,z) = d(u,x) + 1, d(v,z) = d(v,y) + 1, and d(u,y) = d(u,x) + 1, d(v,x) = d(v,y)+ 1, otherwise z ∈[u, v]. Pick a square xxyy. By (PC), one can assume that d(u,x) = d(u,x),d(v,y) = d(v,y). Hence x,y ∈[u, v] and z ∼ x,y. The choice of u, v and (3.1) yield that u ∼ x,x and v ∼ y,y, hence d(u,v) = 3. First suppose that there exists a vertex a ∈[u, y], a x, a = x. Then a ∼ y and a z by (PC). Since d(u,v) = 3,ais not adjacent to v, hence y,y,a,z,v induce a propeller. Thus there exist distinct vertices a,a ∈[u, y] and b,b ∈[x,v], such that a a,b b, x ∼ a,a, and y ∼ b,b.By(PC),y is adjacent to one of the vertices a,a, say y ∼ a. To avoid propellers, we must have a ∼ z, further a ∼ x by (IC4). Similarly, one of the vertices b,b, say b, is adjacent to x ,z,y.Ifa ∼ b , then (x, y) contains a K4 induced by x ,a ,b ,z. Thus a b , whence a ∼ b and b ∼ a by (PC). Since z u, b u, and b z by (PC), the vertices a,x,u,z,b induce a propeller. Case 2. d(u,x) = d(u,y) =: k, d(v,x) = d(v,y) =: m. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.8 (1-18) 8 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

Pick a square xz1yz2. By (PC), either d(u,z1) = k − 1, d(u,z2) = k + 1 and d(v,z2) = m − 1, d(v,z1) = m + 1, or d(u,z1) = d(u,z2) = k and d(v,z1) = d(v,z2) = m. In both cases we obtain z1,z2 ∈[u, v], thus z ∼ z1,z2. In the ﬁrst case we have d(u,z) = k, d(v,z) = m, implying that z ∈[u, v]. Thus only the second possibility may occur. By (3.1) one can ﬁnd vertices u ,v ∼ x,z1 with d(u,u ) = k − 1 and d(v,v ) = m − 1. To avoid a propeller induced by x,z,z1,u,v , the vertex z must be adjacent to one of u ,v , say z ∼ u . Then z v .If u y, then z ∈[u,y], and we are in conditions of Case 1. Thus u ∼ y, hence u = u by the choice of u and v. Consequently, u ∼ z2. Applying (PC) to v and the square xz1yz2 we conclude that either v ∼ y,z2 or v y,z2.Ifv is adjacent to y and z2, then [x,y] violates (IC4) since v ,u,z∈[x,y] however v u, z. Thus v y,z2. Consider a square ysv v . Then d(v,v) = m − 1, d(s,v) = m by (PC). Hence u v, yielding u ∼ s. By (PC) each of the vertices x,z is adjacent to exactly one of the vertices s,v.Ifz ∼ v, then z ∈[u, v] since z ∼ u, v and d(u,v) = d(v,v)+ 2. Thus z ∼ s and z v. On the other hand, if x ∼ v, then u, z, v ∈[x,y] however v z, u, contrary to (IC4). Thus x ∼ s and x v .Ifs = z1, then z1 ∼ s and (x, y) will contain a 4-clique induced by u, z, z1,s.Soz1 = s. Therefore z2 ∼ v by (PC). From (3.1) and the choice of u, v one conclude that v ∼ v,v, whence d(u,v) = 3. Applying (IC4) for [z, v] and (PC), we conclude that there exists a new vertex t ∈[z, v] adjacent to v . Since t is adjacent to at least one of the vertices x,z1, from Case 1 we infer that t ∈[u, v]. Thus t is adjacent to u or v. If t ∼ v, then either t ∼ x,y and the 2-interval [u, t] is not convex, or z ∈[t,x]∪[t,y] and we are in conditions of Case 1. Thus t ∼ u. Then z2 ∼ t and t ∼ x, because the 2-intervals [u, v ] and [z2,v ] are convex. If t z1, then z, u, v ∈[t,z1] and v u, z, contrary to (IC4). Otherwise, if t ∼ z1, then t ∼ y because [x,y] is convex. But in this case (x, y) contains a K4 induced by u, z, z1,t. This concludes the analysis of Case 2, thus establishing the convexity of [u, v]. 2

4. Proof of Theorem 1

We commence by the proof of Theorem 1. (i) ⇒ (ii) and (iii): Let B ⊂ 2[n] be an even -matroid. To show that G(B) satisfies (IC4), pick two bases A,B at distance 2. Applying a twisting, one can suppose that A =∅,B ={1, 2, 3, 4}. Since [A,B] 1 is a 4-octahedron and [A,B]⊆[A,B] 1 , the second half of (IC4) follows. 2 Hn 2 Hn It remains to show that (A, B) contains two non-adjacent bases. By (SEP) one can find a base C ∈ (A, B), say C ={1, 2}. Applying (SEP) to the bases A,B and the elements 1, 2, we will get either two complementary bases C,C of cardinality 2 each (i.e., C ∩ C =∅,C ∪ C = B)or the bases {1,i}, {2,i} for some i ∈{3, 4}, say i = 3. Now, applying (SEP) to A and the element 4, one conclude that {j,4}∈B for some j ∈{1, 2, 3}. Obviously, this base and one of previously defined bases are complementary, establishing the first part of (IC4). Using the induction and axiom (SEP), one can easily show that G(B) is an isometric subgraph 1 of the half-cube 2 Hn. To establish (PC), it suffices to verify this condition for some base-point B and the square A1A2A3A4, where A1 =∅,A2 ={1, 2},A3 ={1, 2, 3, 4}, and A4 ={3, 4} (again twisting does the job). Obviously |B|+|B{1, 2, 3, 4}| = |B{1, 2}| + |B{3, 4}|. Since B 1 + = + the distances d(B,Ai) in G( ) and 2 Hn are equal, we have d(B,A1) d(B,A3) d(B,A2) d(B,A4). This shows that G(B) satisfies the positioning condition (PC). 1 The neighborhood of every vertex of the half-cube 2 Hn is the line graph of the complete graph Kn. Therefore the neighborhood of every vertex A of an arbitrary induced subgraph G of 1 =∅ 2 Hn is the line graph of some subgraph ΓA of Kn:ifsayA , then ΓA consists of all pairs ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.9 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 9 ij such that {i, j} is a vertex of G. So we have established (iii), and (ii) follows from the simple observation that neither W5 nor W6 can occur as an induced subgraph of a line graph. (iii) ⇒ (iv): Assume that a graph G obeys the conditions (IC4) and (PC). Additionally, suppose that there is a vertex b whose neighborhood N(b) is the line graph of a graph Γ = (I, F ) with I ={1, 2,...,n}.LetI ∗ ={1∗, 2∗,...,n∗}. Define the following mapping ϕ : V → 2I . Take b as a base-point and set ϕ(b) =∅. Each vertex x ∈ N(b) encodes some edge ij of Γ ; put ϕ(x) ={i, j}. For a vertex v b, set Cv := N(b)∩[b,v] and ϕ(v) = {ϕ(x): x ∈ Cv}. Clearly ϕ is isotone with respect to the base-point order ≺b: u ≺b v implies ϕ(u) ⊆ ϕ(v). For a vertex v ∈ V, let ϕ∗(v) ={i∗: i/∈ ϕ(v)}. Given an index a ∈ I ∪ I ∗, define ∗ Wa = v ∈ V : a ∈ ϕ(v) ∪ ϕ (v) . = 1 To show that ϕ embeds the graph G (V, E) as an induced subgraph of the half-cube 2 Hn, it suffices to establish that ϕ is (1) injective (i.e., ϕ(u) = ϕ(v) if and only if u = v), and (2) edge- preserving (i.e., |ϕ(u)ϕ(v)|=2 if and only if u ∼ v in G). Since ϕ(b) =∅and G is connected, from (2) would follow that all sets ϕ(v),v ∈ V, have even cardinality. We start with some prop- ∗ ertiesofthemapϕ and sets Wa,a∈ I ∪ I .

(4.1) If x,y ∈ Cv and ϕ(z) ⊆ ϕ(x) ∪ ϕ(y) for z ∈ N(b), then z ∈ Cv.

Proof. Clearly z ∼ x,y. Suppose z/∈[b,v]. Then x ∼ y, because [b,v] is convex. One can assume that ϕ(x) ={1, 2},ϕ(y)={1, 3},ϕ(z)={2, 3}. By (3.1), there exists a common neighbor v0 of x and y one step closer to v. Then z v0, otherwise z ∈[v0,b]⊆[v,b]. Consider a square v0sbt. If it contains x or y, say t = x, then y ∼ s and z s by (PC). Since s ∼ y, the label of s must contain 1 or 3, which is impossible because s is not adjacent to x and z.Sos,t = x,y, therefore x,y ∼ s,t by (IC4). By (PC) one can assume that z ∼ t, z s. But then the vertex t cannot be labelled: on the one hand, ϕ(t) ⊂{1, 2, 3}, however all three 2-subsets of this set have been used already to label the vertices x,y,z. 2

(4.2) The map ϕ : V → 2I is injective.

Proof. The proof is a consequence of the following properties:

(a) for every vertex v b we have [v,b]= conv(Cv); (b) if ϕ(u) ⊆ ϕ(v), then u ≺b v.

To establish (a), proceed by induction on d(v,b).Letwtvs be an upward square with w ≺b v. If d(v,b) = 2, then w = b and s,t ∈ Cv. Since b,v ∈[s,t], the result follows. If d(v,b) > 2, by the induction assumption we have [s,b]=conv(Cs) and [t,b]=conv(Ct ). Since Cs ∪ Ct ⊆ Cv and v ∈[s,t], necessarily v ∈ conv(Cv), whence [v,b]⊆conv(Cv). On the other hand, since [v,b] is convex and Cv ⊂[v,b], we obtain the required equality. To prove (b), suppose that ϕ(u) ⊆ ϕ(v). We assert that Cu ⊆ Cv, with equality if ϕ(u) = ϕ(v). Then (a) would imply that u ≺b v (and u = v if ϕ(u) = ϕ(u)). So, let z ∈ Cu \ Cv. Since ϕ(u) ⊆ ϕ(v), there exist two vertices x,y ∈ Cv such that ϕ(z) ⊆ ϕ(x)∪ ϕ(y). By (4.1) we have z ∈ Cv, contrary to the choice of z. 2

∗ (4.3) For a ∈ I ∪ I , an upward square S = v1v2v3v4 belongs to Wa whenever two opposite vertices of S belong to Wa. In particular, ϕ(v1) ∪ ϕ(v3) = ϕ(v2) ∪ ϕ(v4). ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.10 (1-18) 10 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

∗ Proof. Suppose a is 1 or 1 .Letv1,v3 ∈ Nk(b), v4 ∈ Nk−1(b), and v2 ∈ Nk+1(b). If v2,v4 ∈ W1, then 1 ∈ ϕ(v4) ⊆ ϕ(v1) ∩ ϕ(v3), whence v1,v3 ∈ W1. Similarly, if v2,v4 ∈ W1∗ , then v1,v3 ∈ W1∗ , otherwise 1 ∈ ϕ(v1) ∪ ϕ(v3) ⊆ ϕ(v2). It remains to consider the cases v1,v3 ∈ W1 and v1,v3 ∈ W1∗ . Case 1. v1,v3 ∈ W1. ∈ ∈ ∈ ∈ ∈ ∩ Then 1 ϕ(v2). To show that 1 ϕ(v4), pick x1 Cv1 and x3 Cv3 with 1 ϕ(x1) ϕ(x3).If x1 = x3, then v1,v3 ∈[v2,b],v4 ∈[v1,v3], hence d(x1,v4) = k −2 by convexity of [v2,b]. Thus x1 ∈[v4,b], yielding that 1 ∈ ϕ(v4).Now,letx1 = x3 and d(x1,v3) = d(x3,v1) = k. Applying (PC) to the square S and each of the vertices x1,x3, we deduce that d(x1,v4) = d(x3,v4) = d(b,v4) = k − 1. By (3.1) there exists a vertex x4 ∼ x1,b at distance k − 2tov4. Since [v3,b] is convex and x1 ∈[/ v3,b], necessarily x4 ∼ x3. Suppose 1 ∈/ ϕ(x4). One can assume without loss of generality that ϕ(x1) ={1, 2}, ϕ(x3) ={1, 3}, ϕ(x4) ={2, 3}. Since x1,x4 ∈[b,v1] and ϕ(x3) ⊂ ϕ(x1)∪ϕ(x4), from (4.1) we have x3 ∈[b,v1], which is impossible because d(x3,v1) = k = d(b,v1). Case 2. v1,v3 ∈ W1∗ . Then v4 ∈ W1∗ by deﬁnition of ϕ. Suppose by way of contradiction that 1 ∈ ϕ(v2) and pick ∈ ∈ = − { }= − avertexx2 Cv2 with 1 ϕ(x2).Ifd(x2,v4) k 1, then min d(x2,v1), d(x2,v3) k 1 by (PC), therefore x2 ∈[b,v1]∪[b,v3]. Consequently 1 ∈ ϕ(v1) ∪ ϕ(v3), a contradiction. So d(x2,v4) = k, and from (PC) we infer that d(x2,v1) = d(x2,v3) = k. By (3.1) there exist vertices x1 ∼ x2,b and x3 ∼ x2,b at distance k − 1 from v1 and v3, respectively. If d(x1,v3) = k − 1, then d(x1,v4) = k − 2 by (PC) applied to x1 and v1v2v3v4. This implies d(x2,v4) = k − 1, a contradiction. Thus x1 = x3, moreover x1 ∈[/ b,v3], x3 ∈[/ b,v1].Letϕ(x2) ={1, 2}. Since 1 ∈/ ϕ(x1) ∪ ϕ(x3) and x2 ∼ x1,x3, we conclude that 2 ∈ ϕ(x1) ∩ ϕ(x3), hence x1 ∼ x3.Letϕ(x1) = { } ={ } ∈ ∈ = 2, 3 ,ϕ(x3) 2, 4 . By Case 1 there is a vertex x4 Cv4 with 2 ϕ(x4). Since d(v4,x4) k − 2,x4 is distinct from x1,x2,x3. On the other hand, x4 ∼ x1,x2,x3 because 2 belongs to the labels of all four vertices. Then d(x2,v4) = k − 1, which is impossible. 2

To prove assertions (4.4)–(4.6), we proceed by induction on k = d(b,v).

(4.4) If uv is an upward edge of G and u ≺b v, then |ϕ(v) \ ϕ(u)|=2, and vice versa, if ϕ(u) ⊂ ϕ(v) and |ϕ(v) \ ϕ(u)|=2, then uv is an upward edge of G.

Proof. The result is trivial if u = b. So, let u = b. Pick a neighbor w of u in [u, b] and an upward square wxvy. By the induction assumption and (4.3) we may assume that ϕ(x) = ϕ(w)∪ {1, 2}, ϕ(y) = ϕ(w) ∪{3, 4}, ϕ(v) = ϕ(w) ∪{1, 2, 3, 4}.Ifu coincides with x or y, the result is immediate. Now, let u = x,y, thus u ∼ x,y. Since ϕ(w) ⊂ ϕ(u) ⊂ ϕ(v) and |ϕ(u) \ ϕ(w)|=2, we conclude that ϕ(u) = ϕ(w) ∪{i1,i2} for some i1 ∈{1, 2} and i2 ∈{3, 4}. Conversely, since the map ϕ is isotone, one can easily deduce that |ϕ(v)|=2d(b,v) for every v ∈ V .Now,if ϕ(u) ⊂ ϕ(v), then u ≺b v by assertion (b) in (4.2). In particular, if |ϕ(v) \ ϕ(u)|=2, then v and u lie in consecutive levels, thus u ∼ v. 2

From (4.4) we conclude that ϕ maps the vertices of G to subsets of even cardinality, moreover the images of all vertices in the kth level have cardinality 2k. One can further observe that |ϕ(v)\ ϕ(u)|=2d(v,u) whenever u ≺b v.

(4.5) If i ∈ ϕ(v), then there exists a neighbor x ∈[v,b] of v such that i/∈ ϕ(x). ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.11 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 11

Proof. Pick a neighbor t ∈[v,b] of v. Suppose i ∈ ϕ(t), otherwise is nothing to show. By the induction hypothesis there is a vertex w ∈[t,b], w ∼ t, such that i/∈ ϕ(w). Take an upward square vxwy containing v,w. Since i/∈ ϕ(w), by (4.3) the label of one of the vertices x,y does not contain the index i. 2

(4.6) If u, v ∈ Nk(b), then uv is a (horizontal) edge of G if and only if |ϕ(u) ϕ(v)|=2.

Proof. We may assume that k 2, since the case k = 1 follows from the assumption that N(b) is a line graph. Take u, v ∈ Nk(b) fulﬁlling either one of two reciprocal conditions. If u ∼ v, by (3.1) they have a common parent t. Otherwise, if ϕ(v) = ϕ(u) \{1}∪{2},lett ∈ Nk−1(b) be a neighbor of u such that 1 ∈/ ϕ(t) (its existence follows from (4.5)). Then ϕ(t) ⊂ ϕ(v), thus by (4.4) t is adjacent also to v. In both cases, let w denote a neighbor of t in Nk−2(b). Consider two upward squares S = wxuy and S = wxvy. From (4.3) we know that ϕ(u) = ϕ(x) ∪ ϕ(y) and ϕ(v) = ϕ(x) ∪ ϕ(y). First, assume that u ∼ v.Ift belongs to both S and S, say y = t = y, then (PC) implies that u x,v x and x ∼ x. Therefore ϕ(u) ϕ(v) = ϕ(x)ϕ(x), and we are done because |ϕ(x)ϕ(x)|=2 by the induction hypothesis. Otherwise, if t = x,y, then t ∼ x,y.By (PC), v is adjacent to x or y, say v ∼ x. Applying the induction hypothesis, without loss of generality suppose ϕ(x) = ϕ(w) ∪{1, 2}, ϕ(y) = ϕ(w) ∪{3, 4}, and ϕ(t) = ϕ(w) ∪{1, 3}. By (4.3), ϕ(u) = ϕ(w)∪{1, 2, 3, 4}, while the fact that ϕ is isotone and |ϕ(v)|=|ϕ(u)| implies that ϕ(v) = ϕ(w) ∪{1, 2, 3,i}, for i/∈ ϕ(w) ∪{4}. Hence |ϕ(u) ϕ(v)|=2. As to the converse, assume without loss of generality that ϕ(u) = ϕ(w) ∪{1, 3, 4, 5} and ϕ(v) = ϕ(w)∪{2, 3, 4, 5}, however u v. Since 1, 2 ∈/ ϕ(t), one can suppose that ϕ(t) = ϕ(w)∪ {3, 4}.Ify = t = y, by (4.3) we deduce that ϕ(x) ∩ ϕ(x) = ϕ(w) ∪{5}. Hence x ∼ x, but then the square S and the vertex v violate the positioning condition. Otherwise, if, say, t/∈ S, then t ∼ x,y.If1∈/ ϕ(x) say, then ϕ(x) ⊂ ϕ(v), therefore v ∼ x by (4.4). But then the vertices u, v, w, t, x induce a propeller. 2

Summarizing, from (4.2), (4.3), and (4.6) we deduce that ϕ(G) is an induced subgraph of the 1 ⇒ half-cube 2 Hn, concluding the proof of (iii) (iv). (iv) ⇒ (v): We proceed by induction on the distance k = d(A,B) between A,B in G.If 1 k = 1 or 2 we are done, because G is an induced subgraph of Hn and d 1 (A, B) d(A,B). 2 2 Hn If k 3, take a shortest (A, B)-path in G, (A, C1,...,Ck−2,Ck−1,B). By (IC4) for G there exists a square BC Ck−2C in [B,Ck−2]. Then d(A,C ) = d(A,C ) = k − 1 and by induction 1 = − 1 hypothesis these distances are correct in 2 Hn, as is d(A,Ck−2) k 2. Then by (PC) for 2 Hn applied to A and the square we get d 1 (A, B) = k, as required. 2 Hn (v) ⇒ (i): It sufﬁces to verify (SEP) for sets of the form B =∅and A ={1, 2,...,r}. Further, (SEP) will follow if one can show that for every i ∈ A there exists j ∈ A, j = i, such that A\{i, j} is a vertex of G. We proceed by induction on d(A,B). Since G is an isometric subgraph 1 ∈ = \{ } ∈{ } of 2 Hn, there exists k,l A such that A A k,l is a vertex of G.Ifi k,l , then we are done. Otherwise, since d(A,B)

From the proof of Theorem 1 we infer that the encoding ϕ of G = (V, E) deﬁnes an even -matroid:

Corollary 1. For a graph G = (V, E) satisfying condition (iii) of Theorem 1, the collection Bϕ ={ϕ(v): v ∈ V } is an even -matroid for any base-point b ∈ V and G = G(Bϕ).

Using Theorem 1 and its proof, we obtain the following characterizations of basis graphs of ordinary matroids:

Corollary 2. For a graph G = (V, E) the following conditions are equivalent:

(i) G is a basis graph of a matroid; (ii) [19] G satisﬁes the interval condition (IC3), the positioning condition (PC), and the neighborhood of every vertex is a line graph of a bipartite graph; (iii) G is a basis graph of an even -matroid and the neighborhood of every vertex is a line graph of a bipartite graph.

Proof. The proof of (i) ⇒ (ii) is analogous to the proof of (i) ⇒ (ii) and (iii) of Theorem 1. To establish that (ii) ⇒ (iii), notice that a graph G satisfying condition (ii) of Corollary 2 also satisfies condition (iii) of Theorem 1, thus G is a basis graph of an even -matroid. Finally, assume that G satisfies (IC4), (PC), and the neighborhood N(b) of the base-point b is the line graph of a bipartite graph Γ = (A ∪ B; F). We assert that Bϕ B is a matroid. Evidently, it suffices to show that all the sets ϕ(v) := ϕ(v) B have size k =|B|. We proceed by induction on d(b,v). The assertion is obviously true for b and the vertices in N(b). Otherwise, if v/∈ N(b), pick an upward square uxvy. By induction hypothesis, the sets ϕ(u), ϕ(x), and ϕ(y) have 1 ∅ cardinality k. Then by (PC) for 2 Hn applied to and the square ϕ (u)ϕ (x)ϕ (v)ϕ (y) we get |ϕ(v)|=k, as required. 2

Maurer [19] conjectured that the condition on neighborhoods of vertices in his characterization is redundant. In support of this conjecture, he established [19, Corollary 3.3] that (IC3) and (PC) imply that every neighborhood N(b) is a line graph. In conjunction with our Theorem 1, this leads to the following result:

Corollary 3. Any graph G satisfying the interval condition (IC3) and the positioning condition (PC) is the basis graph of an even -matroid.

Notice that the analogy of Maurer’s conjecture for even -matroids is false: in Section 7, we present two graphs satisfying (IC4) and (PC) in which the neighborhoods of several vertices are not line graphs.

5. Proof of Propositions 1 and 2

Proof of Proposition 1. Let N(b) be the line graph of a graph Γ = (I, F ) having at least two connected components and let Γ1 = (I1,F1) be one such component. Pick b as a base-point and consider the even -matroid Bϕ. Set B1 ={B ∈ Bϕ: B ⊆ I1} and B2 ={B ∈ Bϕ: B ⊆ I \ I1}. The graphs G1 := G(B1) and G2 := G(B2) induce convex subgraphs of the graph G = G(Bϕ): given two vertices A,B of Gi, avertexC of G belongs to [A,B] if and only if A ∩ B ⊆ C ⊆ ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.13 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 13

A ∪ B, hence C belongs to Gi. Therefore both G1 and G2 satisfy condition (IC4), whence from Theorem 1 we infer that B1, B2 are even -matroids and G1,G2 are their basis graphs. To prove that G = G1 G2 it sufﬁces to show that for any two bases B1 ∈ B1 and B2 ∈ B2, the set B1 ∪ B2 is a base of Bϕ. We proceed by induction on |B1|+|B2|. By (SEP) applied to B1, ∅ and B2, ∅ we conclude that there exist i, j ∈ B1 and k,l ∈ B2 such that B := B1 \{i, j}∈B1 1 and B := B2 \{k,l}∈B2. By virtue of induction hypothesis, the sets B0 := B ∪ B = (B1 ∪ 2 1 2 B2) \{i, j, k, l},B := B1 ∪ B = (B1 ∪ B2) \{k,l}, and B := B ∪ B2 = (B1 ∪ B2) \{i, j} are 2 1 bases of Bϕ. By (IC4) applied to Bϕ, there exist two bases A,B which together with B and B form a square of G(Bϕ). Then A ∪ B = B ∪ B = B1 ∪ B2.IfA = B0, then B = B ∪ B , and we are done. So, let A = (B ∪ B ) \{i, k}=(B1 ∪ B2) \{i, k} and B = (B ∪ B ) \{j,l}= (B1 ∪ B2) \{j,l}. Since BB1 ={j}∪(B2 \{l}) and BB ={i}∪(B2 \{l}), the base B 1 has the same distance to the adjacent bases B1 and B1. By triangle condition (3.1) there exists ∈[ ]∩[ ] \{ }= ∩ ⊆ ⊆ ∪ = a base C B,B1 B,B1 adjacent to B1 and B1. Since B1 j B B1 C B B1 (B1 ∪ B2) \{j,l}, we conclude that C = (B1 \{j}) ∪{m} for some m ∈ B2 \{l}. By (SEP) applied to B2, ∅ and element m ∈ B2 ∅, we conclude that there exists m ∈ B2 \{m} such that D := {m, m }∈Bϕ. Since d(C,∅) =|C|=|B1| and d(C,D) =|(B1 \{j}) ∪{m }| = |B1|, by (3.1) there exists a base C ∈[C,∅] ∩ [C,D] adjacent to ∅ and D. Since {m}=C ∩ D ⊆ C ⊆ C, we deduce that C ={e,m} for some e ∈ B1 −{j}. Then e ∈ B1 ⊆ I1 and m ∈ B2 ⊆ I \ I1, therefore the vertices e and m belong to distinct connected components of the graph Γ , contrary to the fact that C ={e,m} is a base of Bϕ. This ﬁnal contradiction shows that B ∪ B = B1 ∪ B2 must be a base, thus establishing that Bϕ = B1 + B2. 2

Proof of Proposition 2. Next suppose B1 and B2 are two even -matroids having the same (unlabelled) indecomposable basis graph G. For a vertex v of G, denote by B1(v) and B2(v) the bases of B1 and B2 represented by v. Pick some base-point b of G. Since the composition of twistings is a twisting and the class of even -matroids is closed with respect to this oper- = =∅ ation, one can assume without loss of generality that B1(b) B2(b) . One can easily see that B1(v) = {B1(x): x ∈ Cv} and B2(v) = {B2(x): x ∈ Cv} for each v b. The collections B ={ ∈ } B ={ ∈ } 1 B1(x): x N(b) and 2 B2(x): x N(b) can be viewed as edge-sets of two labelled connected graphs Γ1 = (I1,F1) and Γ2 = (I2,F2), both having N(b) as a line graph. To Γ1 and Γ2 we will apply the following classical theorem of Whitney [25]: If Γ1 and Γ2 are connected graphs and L(Γ1) L(Γ2), then either Γ1 Γ2 or {Γ1,Γ2}={K3,K1,3}. Further, for any isomorphism β : L(Γ1) → L(Γ2) there exists a unique isomorphism α : Γ1 → Γ2 inducing β, unless Γ1 Γ2 K1,3 +e, Γ1 Γ2 K4 −e, or Γ1 Γ2 K4. As a consequence, if Γ1 and Γ2 do not form an exceptional pair, then any isomorphism between the line graphs L(Γ1) L(Γ2) N(b) is induced by a point-by-point correspondence α of the ground sets I1 and I2. This bijection α maps the bases of B1 to bases of B2 (modulo twisting), and we are done. If {Γ1,Γ2}={K3,K1,3}, then N(b) = K3.IfG contains a ﬁfth vertex a, then obviously a b, thus one can take a at distance 2 from b. Since b and a are contained in a common square, N(b) is not a complete graph, a contradiction. Thus G = K4 and B1 ={∅, {1, 2}, {1, 3}, {1, 4}} and B2 ={∅, {1, 2}, {2, 3}, {1, 3}}. (Every other even -matroid with K4 as a basis graph can be obtained from one of these two by a twisting.) Finally, suppose Γ1 Γ2 ∈{K1,3 +e,K4 −e,K4}. Since B1(b) = B2(b) =∅, |B1(x)|=|B2(x)|=2 for any vertex x of N(b), and |I1|=|I2|=4, the basis graph G may contain only one extra-vertex a, encoded by B1(a) = I1 and B2(a) = I2. Hence G is an induced subgraph of the 4-octahedron, concluding the proof of Proposition 2. 2 ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.14 (1-18) 14 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

6. Proof of Theorem 2

We start with the following characterization of even -matroids (for matroids, an equivalence similar to (i) ⇔ (iii) has been proven by A. Kelmans (unpublished); cf. [24]):

Proposition 3. For B ⊂ 2[n] the following conditions are equivalent:

(i) B is an even -matroid; (ii) [21] for every A,B ∈ B and i ∈ AB,there exists some j ∈ AB,j= i, such that A{i, j}∈B and B{i, j}∈B; (iii) for every A,B ∈ B, there exists i, j ∈ AB, j = i, such that A{i, j}∈B and B{i, j}∈B.

Proof. The equivalence (i) ⇔ (ii) is the main result of [21]. The implication (ii) ⇒ (iii) is trivial. To show that (iii) ⇒ (i), we apply Theorem 1. Obviously, all subsets of B have the same cardinality modulo 2. By induction on d(A,B) one can easily show that G(B) is an isometric (therefore connected) subgraph of the half-cube. It remains to establish (IC4). Let d(A,B) = 2, say B =∅ and A ={1, 2, 3, 4}.Leti = 1 and j = 2 be the indices satisfying the exchange property (iii). Hence B{1, 2}={1, 2} and A{1, 2}={3, 4} are bases. Together with A and B they form a square of G(B). 2

Let B be an even -matroid. Since G(B) ⊆ Π1(B) holds for any collection of even subsets, it remains to establish the converse inclusion. Suppose the segment connecting σ(∅) and σ(B) is an edge of Π1(B), where B ={1, 2, 3, 4,...,k}. By Proposition 3(iii) there exists i, j ∈ B, say i = 1, j = 2, such that both {1, 2} and B \{1, 2} are bases. The point whose first k coordinates are 1 ∅ equal to 2 and the remaining coordinates are 0 is the middle of the segments between σ( ), σ (B) and σ({1, 2}), σ (B \{1, 2}), respectively, contrary to our assumption. Conversely, let B be a collection of sets of even cardinality such that Π1(B) = G(B). Since B 1 G( ) is a connected induced subgraph of 2 Hn, by Theorem 1(iv) it suffices to show that this graph satisfies (IC4). Pick a 2-interval [A,B] of G(B),sayBA={1, 2, 3, 4}. Let H be the [ ] 4-cube induced by the incidence vectors of the subsets of A,B Hn . Suppose by way of contradiction that (A, B) does not contain two non-adjacent bases. Then the incidence vectors of all bases of B included in (A, B) lie in a common facet H of H, say in a facet containing σ(A). Applying a twisting, one can suppose A =∅and B ={1, 2, 3, 4}. Additionally suppose that H is the 3-dimensional subcube of H induced by the incidence vectors of all subsets of B not containing the element 4. We claim that the segment joining σ(A) and σ(B) is an edge of Π(B). Suppose not; then some interior point p = (p1,p2,...,pn) of this segment can be written as a +···+ convex combination μ1σ(B1) μkσ(Bk) of incidence vectors of some bases B1,...,Bk k = = = = = (i.e., each μi is positive and i=1 μi 1). Then p1 p2 p3 p4 λ for some 0 <λ<1 and pj = 0 for any other index j. This implies that no base Bi (i = 1,...,k), contains an element ∈{ } ∈ j/ 1, 2, 3, 4 , whence all Bi are subsets of B.Asp/H, the base B must be one of Bi ,say = = · + k · = = B1 B. Since σ4(Bi) 0fori>1, the equality μ11 i=2 μi 0 λ implies that μ1 λ. · + k = = Replacing μ1 by λ in each of the equalities μ1 1 i=2 μiσj (Bi) λ, j 1, 2, 3, we obtain μi = 0 for all Bi distinct from A and B. Hence the incidence vectors of the bases A,B define an edge of Π1(B), contrary to the initial assumption. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.15 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 15

7. Proof of (ii) ⇒ (iii) of Theorem 1

Let G = (V, E) be a graph satisfying the conditions (IC4) and (PC). To establish that every N(v) (v ∈ V)is a line graph provided it does not contain W5 and W6 as an induced subgraph, we will use the characterization of line graphs due to Beineke (cf. [1] or any other textbook on graph theory): a graph H is a line graph if and only if none of the graphs W5 and F1 − F8 of Fig. 1 is an induced subgraph of H . From (IC4) we infer that induced F2 or F3 cannot occur in the neighborhood of some vertex of G. Each of remaining graphs requires special (and sometimes considerable) efforts. Assume by way of contradiction that N(v) contains an induced subgraph F = Fi,i= 4,...,8, labelled as in Fig. 1. Case F = F1. Suppose N(v) has a K1,3 centered at w and with tips x1,x2,x3. If there exists avertexu ∈ (x1,x2), u = w, not adjacent to one of v or w, say u w, then u ∼ v by (PC) so d(u,x3) 2, thus the square x1ux2w and the vertex x3 provide a counterexample to (PC). Hence, by (IC4) there exist 6 distinct vertices a1,a2 ∈ (x1,x2), b1,b2 ∈ (x1,x3), and c1,c2 ∈ (x2,x3) which are all adjacent to v,w and such that a1 a2,b1 b2,c1 c2. Applying (PC) to these vertices and obtained squares, one can see that each ai is adjacent to one bj and one ck and similar conclusions hold for b-vertices and c-vertices. The subgraph induced by new vertices is triangle-free: if, say, a1,b1,c1 are pairwise adjacent then (c1,x1) contains a 4-clique induced by a1,b1,v,w,contrary to (IC4). Thus this subgraph is an induced 6-cycle, say a1b1c2a2b2c1. Then we have a W6 with center w in N(v), a contradiction. This shows that F1 is forbidden in N(v), thus G does not contain induced propellers. Case F = F4. Since x,y,v ∈ (a1,b1), by (IC4) there is t ∈ (a1,b1) that is not adjacent to exactly one of x,y,v: without loss of generality suppose t v, t ∼ x,y. By (PC) applied to

Fig. 1. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.16 (1-18) 16 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• a2 and the square a1vb1t, we have t a2, and similarly t b2.Nowx,y,t,a2,b2 induce a propeller. Case F = F5. First suppose that [a,b] contains two non-adjacent vertices w1,w2 distinct from v, hence v ∼ w1,w2. To avoid propellers, each of the vertices x,y,z,u must be adjacent to one of w1,w2.Lety ∼ w2. Then w2 ∈[y,b], therefore w2 is adjacent to one of z, u.Ifw2 ∼ z, w2 u, then w2 ∼ x by (PC) applied to x and the square yubw2. But then (z, a) contains a K4 induced by x,y,v,w2.Sow2 ∼ u. Then z w2, otherwise (y, b) would contain a K4 induced by v,z,u,w2. Finally, if x ∼ w2, then [x,b] will be not convex because z, w2 ∈[x,b] and u ∈ [z, w2]\[x,b]. Thus w2 x,z. But then N(v) contains a 5-wheel induced by a,w2,u,z,x,y and centered at y. This shows that there exists a vertex w ∈ (a, b) not adjacent to v.From(PC) we infer that w is not adjacent to any other current vertex. Pick a new vertex s ∈ (y, b). First suppose that s ∼ v. Then s z or s u. In any case, by (PC) we have s ∼ a,w.Now,ifs ∼ u, then s x,z, and in N(y) we obtain a W5 induced by the vertices a,x,z,u,s,v and centered at v. Otherwise, if s ∼ z, s u, by (PC) we conclude that x ∼ s. Then however (z, a) will contain a K4 induced by x,y,s,v. Therefore s v and s ∼ z, u, w, moreover, s is not adjacent to any other current vertex. Analogously, we can ﬁnd a vertex t ∈ (z, a) such that t v, t ∼ w,x,y. Clearly t = s. By (PC) immediately follows that t ∼ s. But then N(y) contains a W5 induced by x,v,t,u,s,z and centered at z. Case F = F6. Pick a new vertex w ∈ (a, b).Ifw ∼ v, then w is adjacent to exactly one of x,y, say w ∼ x, w y. Then w ∼ s,t by (PC), and in N(v) we get a 5-wheel induced by a,x,s,t,b,w and centered at w. Hence any new vertex of (a, b) is not adjacent to v.Picka square S1 = yptp1 containing the vertices y,t.Ifp,p1 are distinct from b,v, then b,v ∼ p,p1 by (IC4). By (PC), a is adjacent to one of p,p1, say a ∼ p. But then p ∈ (a, b) and p ∼ v, contrary to our previous conclusion. Thus one of p,p1, say p1, coincides with b or v.

(a) p1 = b and p ∼ v. Then (PC) implies that p x,w and p ∼ a,s. (b) p1 = v and p ∼ b. Then (PC) implies that p a,s,x and p ∼ w.

Analogously, taking a square S2 = yqsq1 containing the vertices y,s, one may assume that q1 coincides with a or v, leading to conclusions similar to (a) and (b) (by symmetry, one can deﬁne also the squares S3 and S4 for the pairs x,t and x,s, respectively). If the squares S1,S2 are both in position (b), then p ∼ q by (PC) and N(w) will contain a W5 induced by a,x,b,p,q,y and centered at y. Now, suppose that S1 is in position (a) and S2 is in position (b). Again p ∼ q, because p,q,v ∈[y,s] and q v. Then the vertices a,q,p,v,x,b all belong to N(y) and induce a forbidden F5. Finally suppose that all four squares S1,S2,S3,S4 are in position (a). Let S3 = xp tb and S4 = xq sa. Then p,p ∼ a,s,v and q,q ∼ b,t,v, while p,p b,w and q,q a,w. Additionally, p,q x and p,q y. From (PC) we conclude that p ∼ q and p ∼ q. Applying (IC4) to the intervals [b,s] and [a,t], we obtain that p p and q q.But then the vertex w and each of the squares aptp and bqsq provide a counterexample to (PC). Case F = F7. Pick a vertex w ∈ (s, t).Ifw v, then applying (PC) to the square svtw and each of the vertices x and y, we conclude that d(x,w) = d(y,w) = 3. Since a,b ∈[x,w] and y ∈[a,b]\[x,w], we get a contradiction with the convexity of [x,w]. Thus there exist two non- adjacent vertices w1,w2 ∈ (a, b) both adjacent to v. By (PC) each of a,b is adjacent to one of w1,w2. If both a,b are adjacent to the same vertex, say to w1, then x,y,w1 ∈ (a, b). Thus w1 must be adjacent to one of x,y, say w1 ∼ x. Then the square sw1tw2 and the vertex x contradict (PC). Thus let a ∼ w1,a w2 and b ∼ w2,b w1. Then however N(v) contains a forbidden F6 induced by a,x,y,b,t,w1. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.17 (1-18) V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–••• 17

Case F = F8.Picks ∈ (b, c), s = z.First,lets ∼ v.Ifs z, from (PC) we obtain s ∼ x,y and s a. Then (a, s) contains a K4 induced by v,b,x,y, a contradiction. So s ∼ z and then s a, otherwise (b, c) contains K4 induced by v,a,z,s. Now since (s, a) does not have a K4 induced by b,v,x,y, we see that s is not adjacent to one of the vertices x,y, say s x.But then N(v) contains an F5 induced by the vertices x,a,b,z,s,c. Thus any new vertex of [b,c] or [a,c] is not adjacent to v. Take two squares S = bvcp and S = avcq. From (PC) one infer that the vertices p and q are distinct but adjacent and both are adjacent to z. The same condition implies that p a,x,y and q b,x,y. Applying (IC4) for [y,c] and (PC) for q and [y,c], one conclude that there is a vertex s1 ∈ (y, c) adjacent to q.Ifs1 ∼ v, then s1 ∈[v,q]⊆[a,c], contrary to what has been concluded above. Hence s1 v. By (PC) one has p ∼ s1.Now,ifb ∼ s1, then s1 ∈[b,c], implying that s1 and v are adjacent. Thus b s1. Applying (PC) to the square pbys1 and the vertices a,x,z, one conclude that s1 a,x,z. Applying (IC4) for [z, s1] and (PC), one conclude that there exists avertext1 ∈ (z, s1), t1 ∼ y, adjacent to two of the vertices p,c,q. Since t1 ∈[z, y],t1 is also adjacent to exactly two of the vertices a,v,b. Now, if t1 ∼ c, then t1 ∈[c,y], therefore t1 ∼ v. As t1 is adjacent to a or b, we have t1 ∈[a,c]∪[b,c], which is impossible by what has been shown above. So, assume t1 c. From (PC) applied to v and the square t1zcs1 we infer that t1 v. Hence t1 ∼ p,q,a,b, on the other hand, t1 x by (PC) applied to x and the square t1zcs1. Analogously, there exist vertices s2 and t2, such that s2 ∼ x,p,q,c, t2 ∼ x,s2,a,b,z,p,q and t2 y,c,v. If t1 t2, then applying (PC) to c and the square t1at2p we infer that c must be adjacent to t1 or t2. Since this is impossible, we deduce that t1 t2. But then N(z) contains a forbidden F6 induced by b,t1,t2,q,v,c. This contradiction completes the proof of the implication (ii) ⇒ (iii) of Theorem 1. The interval condition (IC4) and the positioning condition (PC) solely do not characterize basis graphs of even -matroids. The following graphs H1 and H2 obey (IC4) and (PC), nevertheless the neighborhoods of several vertices contain induced 5- or 6-wheels. H1 is the graph induced by 11 vertices defined in the analysis of case F = F1. We noticed already that H1 satisfies (IC4) and (PC). Notice also that N(v) contains a 6-wheel induced by the 6-cycle a1b1c2a2b2c1 and centered at w. On the other hand, one can directly check that W5 does not occur in the neighborhoods of vertices of H1. The graph H2 contains 14 vertices x1,...,x4,y1,y2,u1,...,u4,v1,...,v4, where the x-vertices induce a square, the u- and the v-vertices induce two K4, and, additionally, ui and vj are adjacent if and only if i = j. Fur- thermore, y1 ∼ y2 and both y1,y2 are adjacent to all x-vertices. Additionally, y1 is adjacent to all u-vertices and y2 is adjacent to all v-vertices. Finally, the vertices ui,vi are adjacent to the vertices of the edge xi,xi+1 (here indices are taken modulo 4). Using the symmetry and the fact that H2 has diameter 2, one can show that H2 satisfies (IC4) and (PC). On the other hand, N(y2) contains a 5-wheel induced by v1,v2,x3,y1,x1,x2 and centered at x2. Finally notice that W6 does not occur in the neighborhoods of vertices of the graph H2.

Acknowledgments

I am grateful to one of the anonymous referees for a careful reading of earlier and revised versions of the manuscript and numerous corrections and insightful comments that helped to improve the paper. I thank Walter Deuber (postum), Andreas Dress, and Maurice Pouzet, who at different space–time instances pointed me that the classes of graphs I was investigating “depuis longtemps” maybe related to matroids and -matroids. ARTICLE IN PRESS YJCTB:2408 JID:YJCTB AID:2408 /FLA [m1+; v 1.60; Prn:7/06/2006; 16:40] P.18 (1-18) 18 V. Chepoi / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••

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